The New Sudoku Players' Forum

by **bat999** » Fri Jan 09, 2015 5:18 pm

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EDIT
This is a long thread.
My synopsis if you want to 'cut to the chase'.
Here...

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Hi
I've started a new thread because I don't want to clutter Instinct's journal.

Here ---> http://forum.enjoysudoku.com/my-sudoku-learning-journal-t32233.html

Suppose that two bi-value squares A(47) and B(14) are in the same row or column or box.
Work out what happens to an object square X when...
A is not 4
and
B is not 4

If the result in X is the same, then something is known about X because (at least) one of A or B must be not 4.
Square X can now be assigned a value or at least have something eliminated from it.
~~(I think this is known as a 'Kraken' method)~~
So far, so good.

Now with reference to Instinct's post about 'Almost locked sets'...

Suppose that two bi-value squares A(47) and B(14) are in the same row or column or box.
Work out what happens to an object square X when...
A is not 4
and
B is not 4

If the result in X is the not the same, then nothing is known about X, but it indicates that one of the values chosen for A or B were incorrect.
So one of the squares A or B must be 4.
And the 4's can be eliminated from squares that can see both A and B.

Is this true?

Or have I over-presumed.
:lol:

by **Leren** » Fri Jan 09, 2015 8:46 pm

Hi bat, your descriptions are hard to follow as they are unusual by the standards of this (or any other reputable) forum but I'll try to answer them as well as I can.

Firstly your "Kraken" method.

What you have do is ensure that you have covered 100% of possible cases, with a common outcome in your object cell X. If this common outcome is, say, y then you can assign y in X.

What you have to consider in your 2 bi-value squares, is:

(1) A is 4 and B is 4, (2) A is not 4 and B is 4, (3) A is 4 and B is not 4 & (4) A is not 4 and B is not 4 (unless they are the only 2 4's in a row, column or box.)

Since you say that A and B are in the same row, column or box then (1) can't happen and can be ignored. However it looks like you have only considered cases (2) and (3). To correctly apply the method you also have to consider case (4) ie both A and B are not 4 (unless they are the only 2 4's in a row, column or box,which they aren't in instinct's example.)

You have made a similar mistake in the second section.

Your statement: If the result in X is the not the same, then nothing is known about X, but it indicates that one of the values chosen for A or B were incorrect. So one of the squares A or B must be 4. Is not quite right.

What you can conclude is : If the result in X is the not the same, then nothing is known about X, but it indicates that at least one of the values chosen for A or B were incorrect.

So you can't conclude that one of the squares must be 4. They might both be not 4 (unless of course they are the only 2 4's in a row, column or box, which they aren't in instinct's example.)

Now let's have a look at Instinct's Diagram B. What he says here is :

1. r2c6 <> 4 = 1;

2. r2c4 <> 4 = 7 => r2c7 <> 7 = 8 => r1c9 <> 8 = 3 => r1c5 <> 8 = 1; Contradiction (2 1's in Box 2) => at least one of r2c46 must be 4.

This is logically correct, but most experienced solvers regard proof by contradiction as inelegant, if a constructive proof of similar complexity is available.

Fortunately this is true in this case - simply by rearranging things a bit.

A constructive proof is : r2c4 <> 4 = 7 => r2c7 <> 7 = 8 => r1c9 <> 8 = 3 => r1c5 <> 3 = 1 => r2c6 <> 1 = 4.

A more standard notation for this chain is : (4=7) r2c4 - (7=8) r2c7 - (8=3) r1c9 - (3=1) r1c5 - (1=4) r2c6.

If you read the chain from left to right you assume that r2c4 <> 4 and end up concluding that r2c6 = 4.

If you read the chain from right to left you assume that r2c6 <> 4 and end up concluding that r2c4 = 4.

You are then entitled to conclude that at least one of r2c46 must be 4 (They might both be 4 but they can't both be <> 4.) and you have done so without producing a contradiction.

This type of chain is called an XY chain (each of the cells in the chain has 2 digits X and Y with Y in each cell being the same as X in the next cell in the chain, with each cell in the chain sharing a row, column or box with the proceeding cell.

Notice that the "contradiction" part of instinct's argument has been replaced by a linking digit in my XY chain (which says that if r1c5 = 1 then r2c6 <> 1 when read from left to right, or if r2c6 = 1 then r1c5 <> 1 when read from right to left).

Hope this helps.

Leren

by **bat999** » Fri Jan 09, 2015 10:37 pm

Leren wrote:(4) A is not 4 and B is not 4 (unless they are the only 2 4's in a row, column or box.)

Hi
This is not a mistake.
It doesn't matter how many squares in the same row or column or box have the 4 as a candidate.
If you can find any two squares in the same row or column or box that give the same result in the object square when 'off', that's it.
It's not necessary to test any more squares.

It's not clear from the rest of your post whether the answer to my question is 'yes'.

by **Leren** » Fri Jan 09, 2015 11:28 pm

bat 999 wrote:

It doesn't matter how many squares in the same row or column or box have the 4 as a candidate.
If you can find any two squares in the same row or column or box that give the same result in the object square when 'off', that's it.
It's not necessary to test any more squares.

If that's what you continue to believe then good luck !

Here is a recent Kraken move of mine

Code: Select all: *---------------------------------------------------------------------------------* | g9-6dG 249 26 | 49C 7 5 | 3 8 1 | | 5 19 3 | 6 8 19B | 7 4 2 | | 8 14 7 | 134 24 123 | 9 6 5 | |---------------------------+--------------------------+--------------------------| |f 679 5 4 | 2 69 379 | 1 39 8 | | a167cF e39 8 | 147aD 5 17A | 2 39 46bE | |fb169 d239 26 |c1349 469 8 | 5 7 46 | |---------------------------+--------------------------+--------------------------| | 4 8 1 | 79 29 279 | 6 5 3 | | 3 7 5 | 8 1 6 | 4 2 9 | | 2 6 9 | 5 3 4 | 8 1 7 | *---------------------------------------------------------------------------------* Kraken Row 5 Digit 1 : 1 r5c1 - r6c1 = (1-3) r6c4 = r6c2 - (3=9) r5c2 - r46c1 = r1c1 - 6 r1c1; 1 r5c4 - 4 r5c4 = (4-6) r5c9 = r5c1 - 6 r1c1; 1 r5c6 - (1=9) r2c6 - (9=4) r1c4 - r5c4 = (4-6) r5c9 = r5c1 - 6 r1c1; => - 6 r1c1; stte

What you seem to be saying is that was only necessary for me to consider any 2 of the 3 1's in Row 5 to show that r1c1 <> 6 instead of all three 1's.

In cases like this I intend to stick to considering all three 1's.

Leren

by **eleven** » Sat Jan 10, 2015 12:59 am

bat999 wrote:It's not clear from the rest of your post whether the answer to my question is 'yes'.

no

by **bat999** » Sat Jan 10, 2015 1:09 am

eleven wrote:no

OK, thanks.

by **gurth** » Sat Jan 10, 2015 6:30 am

Bat999,
Your first section is correct, only this is NOT the Kraken method, which starts with a WEAK link, in other words a positive assumption rather than the negative "if NOT" of a strong link. I think Leren is describing the conditions necessary for a Kraken. Your initial assumption is that either (NOT one 4) OR (NOT another 4 in same sector) must be true, and this is correct regardless of whether the 4s involved are in bivalue cells or any other cells. E. G. the 2 cells could just as well be 12345 and 234567. If you get the same result for both one 'NOT 4' and the other 'NOT 4', then you CAN deduce that the result is true, no matter what the result may be. That is simple and clear logic. (Bear in mind that very often a wrong assumption can lead to the correct answer for a cell.)

Your second section is incorrect, for the reasons that Leren describes in detail.

by **gurth** » Sat Jan 10, 2015 6:51 am

Leren wrote:This is logically correct, but most experienced solvers regard proof by contradiction as inelegant, if a constructive proof of similar complexity is available.

Leren

This statement is correct logically and any otherwise you can think of, only it is an understatement, most experienced solvers seem to have a phobia about contradictions. Which is absolutely absurd. There is nothing inelegant about contradictions. "Inelegant" is a purely subjective judgement, containing no logic whatsoever. Perhaps these people who hate contradictions so much, have a low tolerance for confrontations and disharmony of any sort.

I will be the first to concede that a proof, for any given case, that is simpler and/or shorter than a contradiction, is more elegant. But not if it is the same length and equally complex.

.

by **blue** » Sat Jan 10, 2015 8:32 am

Hi bat999,

bat999 wrote:Suppose that two bi-value squares A(47) and B(14) are in the same row or column or box.
Work out what happens to an object square X when...
A is not 4
and
B is not 4

If the result in X is the not the same, then nothing is known about X, but it indicates that one of the values chosen for A or B were incorrect.
So one of the squares A or B must be 4.
And the 4's can be eliminated from squares that can see both A and B.

Is this true?

Yes it's true, as long as you're careful amout what is meant by "the result in X is the not the same".
It might be better to think of it as that "the results in X are incompatible".

That would cover cases where:
1) one option forces some value into X, and the other option forces a different value into X.
2) one option forces X to be 'd', and the other eliminates 'd' from X.
3) neither option forces a value into X, but each candidate in X is eliminated for at least one of the options.

(If X is a bivalue cell, then the 3 cases are more or less equivalent).

Note: the logic here, has nothing to do with Almost Locked Sets or that A and B are bi-value cells, and it doesn't require that cells A and B can see each other.

by **bat999** » Sat Jan 10, 2015 10:56 am

@ gurth
Thanks for your reply.
The term 'kraken' was inappropriate for the first part of my question.

@ blue
Thanks for your reply.
Now I seem to have two opposite answers to my question.

by **eleven** » Sat Jan 10, 2015 12:54 pm

Sorry, i had to leave urgently yesterday.
But Gurth and blue already clarified it.
If by "result in X not the same" you mean different numbers, it is true, if you mean different candidates, then not. As long as a common candidate is left in X, both values can be correct.

by **bat999** » Sat Jan 10, 2015 3:10 pm

@ eleven
Thanks for your reply.

gurth said no and blue said yes.
:lol:

This I don't understand:-
If by "result in X not the same" you mean different numbers, it is true, if you mean different candidates, then not.
As long as a common candidate is left in X, both values can be correct.

I will search the www for an explanation and examples.
8-)

by **eleven** » Sat Jan 10, 2015 3:23 pm

What i mean is simply:
Suppose X has candidates 123, and A=4 reduces it to 12, while B=4 reduces it to 23, the "result" for the candidates in X is different, but still both are possible, because X could resolve to 2.

by **bat999** » Sat Jan 10, 2015 4:36 pm

eleven wrote:... the "result" for the candidates in X is different, but still both are possible, because X could resolve...

I've tightened up the question. 8-)

"Different" isn't strong enough.
"Contradictory" nails it.

Suppose that two bi-value squares A(47) and B(14) are in the same row or column or box.
Work out what happens to an object square X when...
A is not 4
and
B is not 4

If the result in X is contradictory, then nothing is known about X, but it indicates that one of the values chosen for A or B were incorrect.
So one of the squares A or B must be 4.
And the 4's can be eliminated from squares that can see both A and B.

Is this true?

PS
I've just read blue's post again.
"It might be better to think of it as that "the results in X are incompatible".
I think me and blue are on the same wavelength.

--------------------------------------------------
These are my two examples of 'contradictory' results.
"The value of square X is 7 when A is not 4. The value of square X is 8 when B is not 4"
and
"The value of square X is 7 when A is not 4. The value of square X is not 7 when B is not 4"
--------------------------------------------------
And this was how blue expressed it...
1) one option forces some value into X, and the other option forces a different value into X.
2) one option forces X to be 'd', and the other eliminates 'd' from X.
But I haven't thought this one through yet.

3) neither option forces a value into X, but each candidate in X is eliminated for at least one of the options.

by **eleven** » Sat Jan 10, 2015 5:48 pm

The New Sudoku Players' Forum

[SOLVED] (not) About 'Almost locked sets'.

[SOLVED] (not) About 'Almost locked sets'.

Re: About 'Almost locked sets'.

Re: About 'Almost locked sets'.

Re: About 'Almost locked sets'.

Re: About 'Almost locked sets'.

Re: About 'Almost locked sets'.

Re: About 'Almost locked sets'.

Re: About 'Almost locked sets'.

Re: About 'Almost locked sets'.

Re: About 'Almost locked sets'.

Re: About 'Almost locked sets'.

Re: About 'Almost locked sets'.

Re: About 'Almost locked sets'.

Re: About 'Almost locked sets'.

Re: About 'Almost locked sets'.